Physics Test - 1

Atoms having the same number of neutrons, but different number of electrons or protons are called Isotones.

Nucleoids having the same atomic number, but a different mass number are known as Isotopes.

Atoms of different chemical elements that have the same number of nucleons are called Isobars.

Conductors \(A O B\) and \(C O D\) are perpendicular to each other shown in figure.

At distance \(a\) above \(O\),

\(B_{1}=\frac{\mu_{0} i_{1}}{2 \pi a}\) 

And \(B_{2}=\frac{\mu_{0} i_{2}}{2 \pi a}\)

\(B_{1}\) is perpendicular to \(B_{2}\).

Resultant of \(B_{1}\) and \(B_{2}\),

\(B=\sqrt{B_{1}^{2}+B_{2}^{2}}\)

\(=\frac{\mu_{0}}{2 \pi a} \sqrt{i_{1}^{2}+i_{2}^{2}}\)

Due to friction between skin and cloths, electrostatic charge is built up on the skin. Hence, electrical discharge may occur when a man touches somebody else. This phenomenon is more significant in winters because due to low humidity, charge has a tendency to stay longer on the body.

The direction of the equipotential surface is from low potential to high potential is not the property of equipotential surfaces.

Any surface over which the electric potential is same everywhere is called an equipotential surface. No work is required to move a charge from one point to another on the equipotential surface. Properties of equipotential surface are:

• The electric field is always perpendicular to an equipotential surface.
• Two equipotential surfaces can never intersect.
• For a point charge, the equipotential surfaces are concentric spherical shells.
• For a uniform electric field, the equipotential surfaces are planes normal to the x-axis.
• The direction of the equipotential surface is from high potential to low potential.

Let \(\mathrm{L}\) is the original length of the wire and \(\mathrm{k}\) is force constant of wire.

Final length \(=\) initial length \(+\) elongation

\(\mathrm{L}^{\prime}=\mathrm{L}+\frac{\mathrm{F}}{\mathrm{k}}\)

For first condition \(\mathrm{a}=\mathrm{L}+\frac{4}{\mathrm{k}} \quad\).....(i)

For second condition \(\mathrm{b}=\mathrm{L}+\frac{5}{\mathrm{k}}\)......(ii)

By solving Eqs. (i) and (ii), we get

\(\mathrm{L}=5 \mathrm{a}-4 \mathrm{~b} \text { and } \mathrm{k}=\frac{1}{\mathrm{~b}-\mathrm{a}}\)

Now, when the longitudinal tension is \(9 \mathrm{~N}\). length of the string.

\(=\mathrm{L}+\frac{9}{\mathrm{k}}=5 \mathrm{a}-4 \mathrm{b}+9(\mathrm{~b}-\mathrm{a}) \)

\(=5 \mathrm{~b}-4 \mathrm{a}\)

The shear is experienced along the surface area of the punch. The surface of the punch is cylindrical with the diameter of 1 cm and a height of 0.3 cm which is the thickness of the sheet.

Therefore force needed to oppose the shear force is = force needed to punch the hole in the steel sheet.

\(F=\) Area \(\times\) Stress

\(F=\) Shear \(\times \pi d t\)

\(F=3.5 \times 10^{8} \times \pi \times 1 \times 10^{-2} \times 0.3 \times 10^{-2}\)

\(F=3.29 \times 10^{4} N \approx 3.3 \times 10^{4} N\)

Given,

\(N_{p}=4 N_{o}\)

\(N_{Q}=N_{o}\)

Also,

\(T_{P}=1 \mathrm{~min}\)

\(T_{Q}=2 \min\)

Now,

Amount left after time

\(N_{P_{t}}=4 N_{O}\left(\frac{1}{2}\right)^{\frac{t}{1}}\)

And

\(N_{Q_{t}}=N_{o}\left(\frac{1}{2}\right)^{\frac{t}{2}}\)

Now,

According to question,

\(N_{P_{t}}=N_{Q_{t}}\)

Thus,

\(4 N_{o}\left(\frac{1}{2}\right)^{t}=N_{o}\left(\frac{1}{2}\right)^{\frac{t}{2}}\)

Then, we get 

\(4=\left(\frac{1}{2}\right)^{\frac{-t}{2}}\)

\(\Rightarrow 4=2^{\frac{t}{2}}\)

Further

\(\frac{t}{2}=2\)

Thus,

\(t=4 \mathrm{~min}\)

Thus,

For \(R\)

\(N_{R}=\left(N_{o}-N_{P_{t}}\right)+\left(N_{o}-N_{Q_{t}}\right)\)

Then,

\(N_{R}=\left(N_{o}-\frac{N_{o}}{4}\right)+\left(N_{o}-\frac{N_{o}}{4}\right)\)

Then, we get

\(N_{R}=\frac{15 N_{o}}{4}+\frac{3 N_{o}}{4}\)

Then,

\(N_{R}=\frac{9 N_{o}}{2}\)

Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave- This statement is incorrect. 

Electromagnetic waves or EM waves: The waves that are formed as a result of vibrations between an electric field and a magnetic field and are perpendicular to each other and to the direction of the wave is called an electromagnetic wave.

Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. They can move in a vacuum.

Properties of electromagnetic waves:

• Not have any charge or we can say that they are neutral.
• Propagate as a transverse wave.
• They move with the velocity the same as that of light i.e., \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).
• It contains energy and they also contain momentum.
• They can travel in a vacuum also.

From above, it is clear that, electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. Therefore, statement in option (A) is correct.

In an electromagnetic wave, the electric field and magnetic field vary continuously with maxima and minima at the same place and same time. Therefore, statement in option (B) is correct.

The energy in an electromagnetic wave is divided equally between electric and magnetic fields. Therefore, statement in option (C) is correct. 

An electromagnetic wave is a perpendicular variation in both the electric field (E) and Magnetic field (B). Therefore, statement in option (D) is incorrect.

In this figure \(Q\) and \(P\) are at the same phase. Therefore, at \(P\) point the path difference between ray \(BP\) and reflected ray \(OP\).

We can say, angles of \(Q O\) and \(O P\) are the same. 

In triangle \(POR , OP =\frac{ PR }{ cos \theta}=\frac{ d }{ cos \theta}\)

In triangle \(QOP,QO = OP \sin \left(90^{\circ}-2 \theta\right)= OP\cos 2 \theta\)

\(\Delta= OP \cos 2 \theta+ OP\)

\(= OP (\cos 2 \theta+1)\)

\(=2 OP \cos ^{2} \theta\)

\(=2 \times \frac{ d }{\cos \theta} \times \cos ^{2} \theta\)

\(=2 d \cos \theta\)

Now, path difference is \(\frac{\lambda}{2}\)

Due to reflection at point \(P\) 

\(\Delta=\frac{\lambda}{2}, \frac{3 \lambda}{2} \ldots \ldots \ldots \ldots \ldots \ldots\)

\(2 d \cos \theta=\frac{\lambda}{2}, \frac{3 \lambda}{2} \ldots \ldots \ldots\)

\(\cos \theta=\frac{\lambda}{4 d }, \frac{3 \lambda}{4 d } \ldots \ldots \ldots \ldots\)

A pressure change at any point in the fluid is transmitted throughout the fluid such that the same change occurs everywhere – Pascal’s Law.

For every action, there is an equal and opposite reaction – Newton’s Third Law

Force is the time rate of change of momentum – Newton’s Second Law

For an ideal gas, the pressure is directly proportional to temperature and constant volume and mass – Ideal Gas Law